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Maru [420]
4 years ago
15

A company produces fruity drinks that contain a percentage of real fruit juice. Drink A contains 5% real fruit juice and Drink B

contains 10% real fruit juice. How many liters of real fruit juice would be needed to produce 250 liters of Drink A and 200 liters of Drink B? How many liters of real fruit juice would be needed to produce aa liters of Drink A and bb liters of Drink B?
Mathematics
1 answer:
Romashka [77]4 years ago
5 0

Answer:

a). 12.5 liters of real fruit juice would be needed to produce 250 liters of Drink A, and 20 liters of real fruit juice would be needed to produce 250 liters of Drink B

b). 0.05aa liters of real fruit juice would be needed to produce aa liters of Drink A, and 0.1bb liters of real fruit juice would be needed to produce bb liters of Drink B  

Step-by-step explanation:

a). Drinks A(250 liters) and Drink B(200 liters)

Total amount of Drink A=250 liters

Real fruit juice=5% of Drink A=(5/100)×250=12.5 liters

12.5 liters of real fruit juice would be needed to produce 250 liters of Drink A

Total amount of Drink B=200 liters

Real fruit juice=10% of Drink B=(10/100)×200=20 liters

20 liters of real fruit juice would be needed to produce 250 liters of Drink B

b). Drinks aa and bb

Total amount of Drink A=aa liters

Real fruit juice=5% of Drink A=(5/100)×aa=0.05aa liters

0.05aa liters of real fruit juice would be needed to produce aa liters of Drink A

Total amount of Drink B =bb liters

Real fruit juice=10% of Drink 10=(10/100)×bb=0.1bb liters

0.1bb liters of real fruit juice would be needed to produce bb liters of Drink B

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Missing Part of Question

d. Refer to part c . If x has a binomial distribution, then so does the number, y , of balls in the sample that meet the USGA’s minimum diameter. [ Note: x + y = 60 ] Describe the distribution of y . In particular, what are p, q , and n ? Also, find E(y) and the standard deviation of y .

Answer:

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d.

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Standard Deviation = 3.098

Step-by-step explanation:

You've answered a and b part of the question, already.

See extracts below

------------------- Answered to Question a and b begins here

"a. What assumptions must be made in order to use the binomial probability distribution to calculate the probability that a particular kind of golf ball will be removed?

A. The experiment consists of n identical trials. There are only two possible outcomes on each trial. The probability of success can change from trial to trial. The trials are dependent.

B. The experiment consists of n identical trials. There are only two possible outcomes on each trial. The probability of success remains the same from trial to trial. The trials are independent.

C. The experiment consists of n identical trials. The number of outcomes can vary. The probability of success can change. The trials are independent.

b. What information must be known in order to use the binomial probability distribution to calculate the probability that a particular kind of golf ball will be removed?

A. The percentage of that kind of golf ball that meets all the requirements

B. The percentage of that kind of golf ball that meets the size requirements

C. The percentage of that kind of golf ball that meets the velocity requirements

D. The percentage of that kind of golf ball that fails to meet both size and velocity requirements

--------------------------- Answer to Question a and b stops here

I'll continue from c and then proceed to d..

c.

Mean is calculated by np

where n = number of observation i.e. sample size = 5 dozens = 5 * 12 = 60

p = probability of success = 20% = 0.2

Mean, E(x) = 0.2 * 60 = 12

Standard Deviation is calculated by the SquareRoot of the products of mean by q (

Standard Deviation = √npq

Where q = 1 - p

q = 1 - 0.2

q = 0.8

Standard Deviation = √(12)(0.8)

Standard Deviation = √9.6

Standard Deviation = 3.098386676965933

Standard Deviation = 3.098 ----- Approximated

d.

From (c) above,

x represents the number of balls that didn't meet USGA minimum diameter

And y represents the number of balls that meet USGA minimum diameter

In total, there are 5 dozen balls

So, x + y = 5 * 12

x + y = 60

So, n = x + y = 60

p, the probability of success of y is the complement of the probability of success of x.

i.e. p of y = 1 - 0.2

p = 0.8

q = 1 - 0.8 = 0.2

Mean is calculated as

E(Y) = np = 60 * 0.8 = 48

Standard Deviation = √npq

Standard Deviation = √(60 * 0.8 * 0.2)

Standard Deviation = √9.6

Standard Deviation = 3.098386676965933

Standard Deviation = 3.098 ----- Approximated

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